24-Pet-A6 Well Logging and Formation Evaluation · May 2017
Question 4 of 8
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2017 – 98-Pet-A6, Reservoir Mechanics (3 hours, open book, non-communicating calculator permitted). The exam consists of EIGHT questions; candidates respond to 100 marks of their choice out of 120. All eight questions are answered in full below.
Reference texts: Craft, B.C. and Hawkins, M. (rev. Terry & Rogers), Applied Petroleum Reservoir Engineering, 3rd ed.; Ahmed, T., Reservoir Engineering Handbook, 5th ed.; Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed.
If doubt exists..." "2. This is an OPEN BOOK EXAM..." "3. The exam consists of EIGHT (8) questions..." "4. Clarity and organization...") immediately above Question 1. every real "Question N" heading on this paper parses cleanly (checked by direct read of the source), so it does not affect the answers below.
Given. A closed (no-influx), undersaturated (single liquid phase) reservoir producing entirely by rock-and-fluid expansion.
Given data
$V_p$
$1\times10^9$ res bbl
$q$
10,000 STB/day
Time produced
6 years
$p_i$
5,000 psi
$c_t$
$1\times10^{-5}$ psi$^{-1}$
$B_o$
1.4 res bbl/STB
Find. The current average reservoir pressure $\bar p$.
Approach. For a closed, undersaturated liquid reservoir, cumulative reservoir-barrel withdrawal is balanced entirely by rock+fluid expansion, giving a direct depletion material balance.
Compressibility-drive material balance. For an undersaturated, volumetric reservoir, $$N_pB_o=c_tV_p\,\Delta p\ \Rightarrow\ \Delta p=\frac{N_pB_o}{c_tV_p}$$
$$\Delta p=\frac{(21.9\times10^6)(1.4)}{(1\times10^{-5})(1\times10^9)}$$
$$\boxed{\Delta p\approx 3{,}066\text{ psi}}$$
Current average reservoir pressure.
$$\bar p=p_i-\Delta p=5{,}000-3{,}066$$
$$\boxed{\bar p\approx 1{,}934\text{ psi}}$$
This large pressure drop for a comparatively modest withdrawal (about 2.2% of the pore volume, in STB terms) illustrates why undersaturated (single-liquid-phase) reservoirs are so pressure-sensitive: with no free gas or aquifer to buffer voidage, every barrel withdrawn must come from a very small total-compressibility expansion.
